Integration by Parts: LIATE, When to Use It, Examples

When integration by parts is the right tool

Integration by parts is the technique for antidifferentiating a product of two functions that come from different families, the kind of integrand where substitution has nothing to grab onto. If you see $x e^x$, $x \sin x$, $x^2 \ln x$, or a lone logarithm or inverse trig function like $\ln x$ or $\arctan x$, parts is almost certainly the move. It is a BC-only topic (CED Topic 6.11), so AB students will not meet it on their exam.

The signature to train your eye on is a product of two factors that pull in opposite directions: one that gets simpler when differentiated (a power of $x$, a logarithm) sitting next to one that stays manageable when integrated (an exponential, a sine, a cosine). Parts is built to exploit exactly that mismatch. A product from the same family, like $\sin x \cos x$, is usually not a parts problem: there a substitution or a trig identity does the job faster.

The quickest way to tell parts from u-substitution is to look for a derivative relationship. Substitution works when one factor is, up to a constant, the derivative of the inside of the other, so the whole integrand collapses onto a single new variable. In $\int 2x e^{x^2}\,dx$ the factor $2x$ is the derivative of $x^2$, so substitution wins. In $\int x e^x\,dx$ the factor $x$ has nothing to do with the derivative of $e^x$, substitution stalls, and parts takes over. Always test substitution first; only when it stalls on a genuine product do you reach for parts.

The formula and why it works

Integration by parts is the product rule read backward. The product rule says $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$. Multiply through by $dx$ to reach the differential form $d(uv) = u\,dv + v\,du$, then integrate both sides to get $uv = \int u\,dv + \int v\,du$. Solving for the piece you want gives the formula.

That single line is the whole method. You split the integrand into a part you name $u$ and a part you name $dv$, then trade the original integral $\int u\,dv$ for a new one, $\int v\,du$. The trade is only worth making if $\int v\,du$ turns out easier than the integral you started with. Nothing guarantees that: it depends entirely on how you split the integrand, which is what the next section is about.

LIATE: choosing u without guessing

When the integrand is a product, LIATE gives you a priority order for what to call $u$. Read down the list; whichever type appears first becomes $u$, and everything else, including the $dx$, becomes $dv$.

1. Logarithmic, such as $\ln x$ and $\log x$
2. Inverse trigonometric, such as $\arctan x$ and $\arcsin x$
3. Algebraic, meaning powers of $x$ like $x$ and $x^2$
4. Trigonometric, such as $\sin x$ and $\cos x$
5. Exponential, such as $e^x$

The ordering is not arbitrary. Functions near the top get simpler when you differentiate them, and differentiating $u$ is exactly what the formula does to it: $\ln x$ becomes $\frac{1}{x}$, and a power of $x$ drops a degree toward a constant. Functions near the bottom stay well behaved when you integrate them, and integrating $dv$ is what the formula does to that piece: $e^x$ integrates to $e^x$, and $\sin x$ integrates to $-\cos x$. Putting the top function in $u$ and the bottom function in $dv$ steers $\int v\,du$ toward something simpler than where you began.

Treat LIATE as a heuristic, not a law. It is a reliable opening guess that works on nearly every AP problem, but it is not a theorem and it does not know anything about your specific integrand. The genuine test of a good split is whether $\int v\,du$ actually came out easier. If your choice makes the new integral harder than the original, you picked $u$ and $dv$ backward: swap them and run the formula again.

Pitfalls and the cases LIATE hides

Three mistakes account for most lost points. First, the minus sign: the formula is $uv - \int v\,du$, and dropping that subtraction quietly wrecks the answer. Second, $dv$ has to be something you can actually integrate to recover $v$; if you cannot antidifferentiate your chosen $dv$, that is a signal you split the integrand wrong. Third, this is still an indefinite integral, so the constant $+C$ belongs at the end.

A single logarithm or inverse trig function looks like it has no $dv$ at all, but it does: let $dv = dx$, which makes $v = x$. That is how $\int \ln x\,dx$ and $\int \arctan x\,dx$ get done, and it is the case LIATE quietly relies on, since Logarithmic and Inverse trig sit at the top of the list and $dx$ is the only thing left to be $dv$.

Some integrands need parts more than once. A polynomial of degree $n$ multiplied by $e^x$ or $\sin x$ takes $n$ applications, each one knocking the power down by a degree, and the tabular method is just an organized way to track that repetition. A handful of integrals such as $\int e^x \sin x\,dx$ return to themselves after two rounds of parts; when the original integral reappears, set it equal to $I$ and solve the resulting equation for $I$ algebraically instead of integrating forever. For a definite integral, the boundary term gets evaluated too:

Check by differentiating

Integration by parts produces answers with several terms, and a stray sign or dropped factor is easy to miss, so build the habit of differentiating your result and confirming it collapses back to the integrand. Each worked example on this page ends that way. Because you are differentiating a product, the product rule you started from reappears and the terms telescope: the derivative of $uv$ contributes $u\,dv + v\,du$, the $-\int v\,du$ term contributes $-v\,du$, and those two $v\,du$ pieces cancel, leaving exactly $u\,dv$. If your check does not reduce to the original integrand, the error is almost always a sign or a missed factor in $v$.